Question

A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m . Find the worked done by (a) the 255 N force and (b) friction. Assume that the coefficient of kinetic friction is 0.32

Answer 1

(a) By the definition of the work done, we have:

$$W=Fdcos\theta=225\ N\times7.5\ m\times cos60^{\circ}=843.75\ J.$$

(b) Let's first find the force of friction:

$$F_{fr}=-\mu_kN=-\mu_k(mg+Fsin\theta).$$

Finally, we can find the work done by the force of friction:

$$W=-\mu_kd(mg+Fsin\theta),$$ $$W=-0.32\times7.5\ m\times(25\ kg\times9.8\ \dfrac{m}{s^2}+225\ N\times sin60^{\circ}),$$ $$W=-1056\ J.$$

The sign minus means that the force of friction does negative work.